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Tuesday, 27 December 2011

Series-6 (CAT-2004)

Consider the sequence of numbers a1, a2,
a3,... to infinity, where a1 = 81.33 and a2 =
-19 and aj = aj-1 – aj-2. What is the sum of
first 6002 terms of this sequence?(1)-100.33 (2)-30.00 (3)62.33 (4)119.33Solution
follows here:

Solution:

aj= aj-1– aj-2=> a3 = a2
– a1;

a4 = a3 – a2 = a2
– a1 – a2 = -a1;

a5 = a4 – a3 = -a1
– (a2 – a1)= -a2;

a6 = a5 – a4 = -a2
– (– a1)= a1-a2;

a7 = a6 – a5 = a1-a2
– (–a2)= a1;

As a7 is being equal to a1,
from a7 onwards, the series is starting to repeat.

=> First six terms are getting repeated

a1+a2+a3+a4+a5+a6
= a1+a2+( a2 – a1)+( -a1)+(
-a2)+( a1-a2) = 0

Sum of first six terms is zero => sum of next
six terms is also zero…

So sum of first 6000 terms is also zero, as 6000
is a multiple of six.

Sum
of first 6002 terms = (sum of first 6000 terms)+( sum of first 2 terms)